A boat moving at 15 km/h relative to water is crossing a river 1.5 km wide. The river flows at 5.0 km/h
a) if the boat heads directly for the opposite shore, how long will the trip take?
b)Where will the boat arrive on the opposite shore?
c)If the boat is to reach a point directly across from the starting point , in which direction should the boat head?
d)How long will the crossing take in case c)?
It will take 1.5/15 hr to get across
IT does downstream 5*1.5/15 km
For the c), draw the triangle. On side is 5km/hr, the hypotenuse is 15km/hr. Solve for the side going straight across, that is the velocity across. For the angle, SinTheta=5/15
i didn't get the answer of bob very well. can he help me more widely.
the question is
Q)A boat moving at 15 km/h relative to water is crossing a river 1.5 km wide. The river flows at 5.0 km/h
a) if the boat heads directly for the opposite shore, how long will the trip take?
b)Where will the boat arrive on the opposite shore?
c)If the boat is to reach a point directly across from the starting point , in which direction should the boat head?
d)How long will the crossing take in case c)?
use this, Vx = dx * t
then use that t and find
Vy = dy * t
x and y are x dimension n y dimension
uhh..
i'll just do it,
Vx= 15 km/hr
dx = 1.5 km
dy = 5 km/hr
To answer these questions, we need to break down the boat's motion into its horizontal and vertical components. Let's use the following notations:
v_b: velocity of the boat relative to water
v_r: velocity of the river
d: width of the river
θ: angle between the boat's heading and the vertical direction
a) To find the time it takes for the boat to cross the river, we need to consider the horizontal component of the boat's motion. The boat's velocity relative to the ground can be found using vector addition:
v_gb = v_b + v_r
Since the boat is heading directly for the opposite shore, the angle between the boat's heading and the vertical direction is 90 degrees. So the horizontal component of the boat's velocity can be found using trigonometry:
v_horizontal = v_gb * cos(θ)
In this case, θ = 90 degrees, so cos(θ) = 0.
Therefore, the horizontal component of the boat's velocity is 0 km/h. This means the boat will only be moving vertically due to the river's current.
To find the time it takes to cross the river, we can use the formula:
time = distance / velocity
In this case, since the horizontal component of the velocity is 0 km/h, the boat will not cover any horizontal distance. So the time it takes to cross the river is 0.
Therefore, the answer to (a) is that the trip will take 0 hours.
b) Since the boat is not moving horizontally, it will arrive directly across from its starting point. The vertical component of its position will change due to the river's current, but the horizontal component will remain the same.
Therefore, the boat will arrive at the same horizontal position on the opposite shore as its starting point.
c) To reach a point directly across from the starting point, the boat needs to counteract the river's current and move in the opposite direction. This means the boat should head upstream.
d) In case (c), the boat's velocity relative to the ground will be equal to the river's velocity, but in the opposite direction:
v_gb = -v_r
Using the same formula as in (a), we can find the time it takes to cross the river:
time = distance / velocity
In this case, the distance is still the width of the river, d = 1.5 km, and the velocity is the magnitude of the boat's velocity relative to the ground, which is equal to the magnitude of the river's velocity:
velocity = |v_gb| = |v_r|
Therefore, the answer to (d) is that the crossing will take 1.5 km / 5.0 km/h = 0.3 hours, or 18 minutes.